Free CAS MAS-I (Modern Actuarial Statistics I) Statistics Practice Questions

Statistical inference on CAS Exam MAS-I covers hypothesis testing, confidence interval construction, Bayesian estimation methods, Monte Carlo simulation, and applied statistics for P&C actuarial problems (CAS).

182 Questions
55 Easy
70 Medium
57 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
An actuary records five independent claim severities (in thousands):

8,  12,  15,  18,  228, \; 12, \; 15, \; 18, \; 22

Calculate the unbiased sample variance.
Solution
E is correct. First compute the sample mean: Xˉ=8+12+15+18+225=755=15.\bar{X} = \frac{8 + 12 + 15 + 18 + 22}{5} = \frac{75}{5} = 15. The deviations from the mean are 7,3,0,3,7-7, -3, 0, 3, 7, with squared deviations 49,9,0,9,4949, 9, 0, 9, 49, summing to 116116. The unbiased sample variance uses the divisor n1=4n - 1 = 4: S2=(XiXˉ)2n1=1164=29.S^2 = \frac{\sum (X_i - \bar{X})^2}{n - 1} = \frac{116}{4} = 29. The value 29 falls in the interval [25,30)[25, 30), so the answer is E.
Question 2 Medium
Let X1,X2,X3,X4,X5X_1, X_2, X_3, X_4, X_5 be a random sample of five independent observations from a continuous Uniform(0,10)(0, 10) distribution. Calculate the probability that the sample maximum X(5)X_{(5)} exceeds 8.
Solution
B is correct. The maximum X(5)X_{(5)} is at most 8 if and only if every observation is at most 8. By independence, P(X(5)8)=i=15P(Xi8)=[FX(8)]5=(8/10)5=0.85=0.32768.P(X_{(5)} \leq 8) = \prod_{i=1}^{5} P(X_i \leq 8) = [F_X(8)]^5 = (8/10)^5 = 0.8^5 = 0.32768. Therefore P(X(5)>8)=10.32768=0.67232P(X_{(5)} > 8) = 1 - 0.32768 = 0.67232. This value falls in the interval [0.65,0.80)[0.65, 0.80).
Question 3 Hard
An actuary observes six iid claim amounts drawn from an exponential distribution with mean θ=250\theta = 250. Let X(1)X(2)X(6)X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(6)} denote the order statistics.

Calculate P(X(3)>200).P(X_{(3)} > 200).
Solution
D is correct.

Step 1: Translate the event in terms of how many observations fall on each side of 200. X(3)>200X_{(3)} > 200 means the third-smallest of the six exceeds 200, equivalently at most two of the six observations are 200.\leq 200.

Step 2: Define a binomial counter. Let Y=#{i:Xi200}Y = \#\{i : X_i \leq 200\}. Since the XiX_i are iid, YBinomial(6,p)Y \sim \text{Binomial}(6, p) with p=P(X200).p = P(X \leq 200).

Step 3: Compute pp using the exponential cdf.
p=1e200/250=1e0.810.4493=0.5507,1p0.4493.p = 1 - e^{-200/250} = 1 - e^{-0.8} \approx 1 - 0.4493 = 0.5507,\quad 1 - p \approx 0.4493.

Step 4: Sum the binomial probabilities for Y2Y \leq 2.
P(Y=0)=(0.4493)60.00823P(Y = 0) = (0.4493)^6 \approx 0.00823
P(Y=1)=6(0.5507)(0.4493)560.55070.018310.06051P(Y = 1) = 6(0.5507)(0.4493)^5 \approx 6 \cdot 0.5507 \cdot 0.01831 \approx 0.06051
P(Y=2)=15(0.5507)2(0.4493)4150.303270.040750.18539.P(Y = 2) = 15(0.5507)^2(0.4493)^4 \approx 15 \cdot 0.30327 \cdot 0.04075 \approx 0.18539.

Step 5: Add.
P(X(3)>200)=P(Y2)0.00823+0.06051+0.185390.254.P(X_{(3)} > 200) = P(Y \leq 2) \approx 0.00823 + 0.06051 + 0.18539 \approx 0.254.

The single-observation tail P(X>200)=e0.80.449P(X > 200) = e^{-0.8} \approx 0.449 is a tempting shortcut but ignores the order-statistic structure entirely; it answers a different question.

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